OE: Practice SET-1
GRE Math: Logic & Integer Properties
1.
If \(x\) and \(y\) are integers and \(3x + 4y\) is odd, which of the following must be true?
Solution (A): Since \(y\) is an integer, \(4y\) must be even.
The expression is \(3x + (\text{Even}) = \text{Odd}\).
This means \(3x\) must be odd.
For \(3x\) to be odd, \(x\) must be odd (since \(\text{Odd} \times \text{Odd} = \text{Odd}\)).
2.
If \(c\) and \(d\) are integers and \(c^2 – d^2\) is odd, which of the following must be true?
Solution (D): If \(c^2 – d^2\) is odd, then \(c^2\) and \(d^2\) must have different parity (one is even, one is odd).
If \(c^2\) is even, \(c\) is even. If \(c^2\) is odd, \(c\) is odd.
Therefore, \(c\) and \(d\) must have different parity.
If one is even and one is odd, their sum \(c + d\) must be odd.
3.
If \(p\), \(q\), and \(r\) are prime numbers and \(p = q + r\), which of the following must be true?
Solution (C): If \(q\) and \(r\) are both odd primes, their sum \(p\) would be \( \text{Odd} + \text{Odd} = \text{Even} \).
If \(p\) is an even prime, \(p\) must be 2. But \(p = q+r\) and \(q, r \ge 3\), so \(p \ge 6\). This is a contradiction.
The only way this works is if one of the primes being added (\(q\) or \(r\)) is the even prime, 2.
Example: \(5 = 3 + 2\).
4.
Define a function \( [n] \) as \( [n] = n/2 \) if \(n\) is even, and \( [n] = n+1 \) if \(n\) is odd. What is the value of \( [7] + [10] \)?
Solution (B):
1. Evaluate \( [7] \): 7 is odd, so use \(n+1 \to 7+1 = 8\).
2. Evaluate \( [10] \): 10 is even, so use \(n/2 \to 10/2 = 5\).
3. Sum the results: \(8 + 5 = 13\).
5.
If \(x\) and \(y\) are both prime numbers greater than 2, which of the following expressions must be even?
Solution (A): If \(x\) and \(y\) are prime numbers greater than 2, they must both be odd.
A) \(x + y = \text{Odd} + \text{Odd} = \text{Even}\). This must be true.
B) \(xy = \text{Odd} \times \text{Odd} = \text{Odd}\).
C) \(x – y + 1 = \text{Odd} – \text{Odd} + \text{Odd} = \text{Even} + \text{Odd} = \text{Odd}\).
D) \(y + 2 = \text{Odd} + \text{Even} = \text{Odd}\).
E) \(x^y = (\text{Odd})^{\text{Odd}} = \text{Odd}\).
6.
A two-digit number \(K\) is created. A new number \(L\) is formed by reversing the digits of \(K\). If \(K + L\) is a perfect square, which of the following could be \(K\)?
Solution (C): Let \(K = 10a + b\). Then \(L = 10b + a\).
The sum \(K + L = (10a + b) + (10b + a) = 11a + 11b = 11(a+b)\).
For this sum to be a perfect square, \((a+b)\) must be 11 (or 44, etc., but 11 is the only one possible for two digits).
We test the options to see which has digits that sum to 11.
C) 29: \(2 + 9 = 11\).
Check: \(K=29, L=92\). \(K+L = 29+92 = 121\), which is \(11^2\).
7.
The sum of the first 20 positive even integers is 420. What is the sum of the even integers from 42 to 80, inclusive?
Solution (E): This is an arithmetic series. The info about the first 20 is extra.
1. Find the number of terms \(n\): \( \frac{(\text{Last} – \text{First})}{\text{Step}} + 1 = \frac{(80 – 42)}{2} + 1 = \frac{38}{2} + 1 = 19 + 1 = 20 \) terms.
2. Find the average: \( \frac{(\text{First} + \text{Last})}{2} = \frac{(42 + 80)}{2} = \frac{122}{2} = 61 \).
3. Find the sum: \( \text{Sum} = n \times \text{Average} = 20 \times 61 = 1220 \).
8.
If \(a\) is an odd integer and \(b\) is an even integer, which of the following expressions must be odd?
Solution (B):
A) \(b^a = (\text{Even})^{\text{Odd}} = \text{Even}\).
B) \(a – b = \text{Odd} – \text{Even} = \text{Odd}\). This must be true.
C) \(ab = \text{Odd} \times \text{Even} = \text{Even}\).
D) \(2a + b = \text{Even} + \text{Even} = \text{Even}\).
E) \(a^b = (\text{Odd})^{\text{Even}} = \text{Odd}\). Wait, both B and E are odd.
*Correction:* Let’s re-check exponents. \(a=3, b=2\). \(a^b = 3^2 = 9\) (Odd).
\(a=3, b=0\) (0 is even). \(a^b = 3^0 = 1\) (Odd). So E is always odd.
\(a-b\). \(3-2 = 1\) (Odd). \(3-0 = 3\) (Odd).
This question is flawed as B and E are both “must be odd”. Let me fix option E.
New E: \(a+b+1\). \(\text{Odd} + \text{Even} + \text{Odd} = \text{Odd} + \text{Odd} = \text{Even}\).
8.
If \(a\) is an odd integer and \(b\) is an even integer, which of the following expressions must be odd?
Solution (B):
A) \(b^a = (\text{Even})^{\text{Odd}} = \text{Even}\) (assuming \(b \ne 0\)).
B) \(a – b = \text{Odd} – \text{Even} = \text{Odd}\). This must be true.
C) \(ab = \text{Odd} \times \text{Even} = \text{Even}\).
D) \(2a + b = \text{Even} + \text{Even} = \text{Even}\).
E) \(a + b + 1 = (\text{Odd} + \text{Even}) + \text{Odd} = \text{Odd} + \text{Odd} = \text{Even}\).
9.
If \(x\) is a multiple of 3 and \(y\) is a multiple of 6, which of the following must be true?
Solution (C):
\(x = 3k\) for some integer \(k\).
\(y = 6j\) for some integer \(j\).
A) \(x=3, y=6\). \(x+y=9\). Not a multiple of 6.
C) \(xy = (3k)(6j) = 18(kj)\). Since \(k\) and \(j\) are integers, \(kj\) is an integer. Thus \(xy\) must be a multiple of 18.
D) \(x=3, y=6\). \(y/x=2\). But if \(x=9, y=6\), \(y/x\) is not an integer.
10.
If \(k\) is a positive integer, which of the following expressions *cannot* be an even integer?
Solution (E):
\(2k\) must be even.
\(2k – 1 = (\text{Even}) – 1 = \text{Odd}\).
This expression is always odd, so it can never be an even integer.
A) If \(k=3\), \(k-1=2\) (Even).
C) \(k(k+1)\) is the product of consecutive integers, which is always even.
D) If \(k=2\), \(k^2=4\) (Even).
Score: 0 / 10